Find Minimum in Rotated Sorted Array


Problem Statement :


Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

[4,5,6,7,0,1,2] if it was rotated 4 times.
[0,1,2,4,5,6,7] if it was rotated 7 times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

 

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 
 

Constraints:

n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
All the integers of nums are unique.
nums is sorted and rotated between 1 and n times.


Solution :



title-img 


                            Solution in C :

int findMin(int* nums, int numsSize){
    int l = 0, r = numsSize - 1;
    while (l < r) {
        int m = l + (r - l) / 2;
        if (nums[m] > nums[r]) l = m + 1;
        else r = m;
    }
    return nums[l];
}
                        


                        Solution in C++ :

class Solution {
public:
    int findMin(vector<int>& nums) {
        int n = nums.size();
        int low=0, high=n-1;
        
        while(low<high){
            if(nums[low] <= nums[high]) return nums[low];
            int mid = low + (high-low)/2;
            if(nums[low] > nums[mid]){
                high=mid;
            } else if(nums[mid] > nums[high]) {
                low=mid+1;
            } 
        }
        if(nums[low] <= nums[high]) return nums[low];
        return -1;
    }
};
                    


                        Solution in Java :

class Solution {
  public int findMin(int[] nums) {
    int l = 0;
    int r = nums.length - 1;

    while (l < r) {
      final int m = (l + r) / 2;
      if (nums[m] < nums[r])
        r = m;
      else
        l = m + 1;
    }

    return nums[l];
  }
}
                    


                        Solution in Python : 
                            
class Solution:
    def findMin(self, nums: List[int]) -> int:
        l, r = 0, len(nums) - 1

        while l < r:
            m = l + (r - l)

            if nums[m] > nums[r]:
                l = m + 1
             
            else:
                 r = m 

        return nums[l]


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